Question

1. an archer pulls her bowstring back 0.400 m by exerting a force that increases uniformly from zero to 230 n. (a) what is the equivalent spring constant of the bow? (b) how much work does the archer do on the string in drawing the bow?

Explanation:

Part (a)

Step1: Recall Hooke's Law

Hooke's Law states that \( F = kx \), where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the displacement. The force increases uniformly from 0 to 230 N, so the average force is used, but for the spring constant, we can consider the maximum force at maximum displacement. At maximum displacement \( x = 0.400 \, \text{m} \), the force \( F = 230 \, \text{N} \) (since it increases uniformly to that value).

Step2: Solve for \( k \)

From \( F = kx \), we can rearrange to \( k=\frac{F}{x} \). Substituting \( F = 230 \, \text{N} \) and \( x = 0.400 \, \text{m} \), we get \( k=\frac{230}{0.400}= 575 \, \text{N/m} \).

Part (b)

Step1: Recall Work Done on a Spring

The work done on a spring (or a system with a variable force like this, where force increases uniformly) is given by the formula for the work done by a variable force, which for a linear force (like Hooke's Law) is the area under the force - displacement graph. Since the force increases from 0 to \( F \) over displacement \( x \), the graph is a triangle, and the work \( W=\frac{1}{2}Fx \).

Step2: Calculate the Work

We know \( F = 230 \, \text{N} \) and \( x = 0.400 \, \text{m} \). Substituting into the formula \( W=\frac{1}{2}\times230\times0.400 \). First, calculate \( 230\times0.400 = 92 \), then \( \frac{1}{2}\times92 = 46 \, \text{J} \).

Part (a) Answer:

The spring constant \( k=\boxed{575 \, \text{N/m}} \)

Part (b) Answer:

The work done is \( \boxed{46 \, \text{J}} \)

Answer:

Step1: Recall Work Done on a Spring

The work done on a spring (or a system with a variable force like this, where force increases uniformly) is given by the formula for the work done by a variable force, which for a linear force (like Hooke's Law) is the area under the force - displacement graph. Since the force increases from 0 to \( F \) over displacement \( x \), the graph is a triangle, and the work \( W=\frac{1}{2}Fx \).

Step2: Calculate the Work

We know \( F = 230 \, \text{N} \) and \( x = 0.400 \, \text{m} \). Substituting into the formula \( W=\frac{1}{2}\times230\times0.400 \). First, calculate \( 230\times0.400 = 92 \), then \( \frac{1}{2}\times92 = 46 \, \text{J} \).

Part (a) Answer:

The spring constant \( k=\boxed{575 \, \text{N/m}} \)

Part (b) Answer:

The work done is \( \boxed{46 \, \text{J}} \)